A gas expands and does 82.7 J of work on the surroundings. At the same time, the gas loses 103 J of energy to the surroundings as heat. What is the change in the internal energy of the gas? would this be ΔE = q + w (-103) + (82.7) = -20.3. I get confused with the positive and negative charges.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Ignore the heat part first. If the gas does work on its surroundings, do you think its energy increases or decreases? If you push a block of metal, does your energy increase or decrease? $\endgroup$– orthocresolOct 16, 2015 at 21:37
-
1$\begingroup$ This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. $\endgroup$– bonOct 17, 2015 at 10:23
-
$\begingroup$ Related: Thermodynamics, pressure volume work $\endgroup$– user7951Nov 12, 2015 at 16:17
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is
$$\Delta U = Q + W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur).
Therefore, heat transferred from the surroundings to the system is assigned a positive sign in the equation, whereas heat transferred from the system to the surroundings is assigned a negative sign.
Similarly, work done by the surroundings on the system is assigned a positive sign in the equation, whereas work done by the system on the surroundings is assigned a negative sign.
(Note that some books on physics or engineering still use a non-standard definition when discussing the work done by a thermodynamic heat engine.)